Question

CH4 , NH3 , and H2O have same hybridisation but different geometries and bond angles . Explain the geometries and bond angles of all the molecules and give reason for the same .

Answer 1

It is because of the difference in lone pairs.
CH4 has no lone pair.[Tetrahedral]
NH3 has 1 lone pair.[Pyramidal]
H2O has 2 lone pair.[Bent]

Regards,
Adithiya.

Answer 2

Answer:

$\text{Number of electrons}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

First we have to determine the hybridization of the following molecules.

(a) $CH_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electrons is 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral and the bond angle will be, $109.5^o$

(b) $NH_3$

$\text{Number of electrons}=\frac{1}{2}\times [5+3]=4$

The number of electrons is 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are three atoms around the central nitrogen, the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be trigonal pyramidal and the bond angle will be, $107^o$

(c) $H_2O$

$\text{Number of electrons}=\frac{1}{2}\times [6+2]=4$

The number of electrons is 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are two atoms around the central oxygen, the third and fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent and the bond angle will be, $104^o$.